Suppose a current I is flowing around a loop, when someone suddenly cuts the wire. The current drops “instantaneously” to zero. This generates a whopping back emf, for although I may be small, d I/dt is enormous. (That’s why you sometimes draw a spark when you unplug an iron or toaster—electromagnetic induction is desperately trying to keep the current going, even if it has to jump the gap in the circuit.)
Nothing so dramatic occurs when you plug in a toaster or iron. In this case induction opposes the sudden increase in current, prescribing instead a smooth and continuous buildup. Suppose, for instance, that a battery (which supplies a constant emf \varepsilon_{0}) is connected to a circuit of resistance R and inductance L (Fig. 7.35). What current flows?
Question 7.12: Suppose a current I is flowing around a loop, when someone s...
The total emf in this circuit is \varepsilon_{0} from the battery plus −L(d I/dt) from the inductance.Ohm’s law, then, says ^{17}
\varepsilon_{0}-L\frac{dI}{dt}=IR.This is a first-order differential equation for I as a function of time. The general solution, as you can show for yourself, is
I(t)=\frac{\varepsilon _{0}}{R}+ke^{-(R/L)t},where k is a constant to be determined by the initial conditions. In particular, if you close the switch at time t = 0, so I (0) = 0, then k = −\varepsilon _{0}/R, and
I(t)=\frac{\varepsilon _{0}}{R}\left[1-e^{-(R/L)t}\right] . (7.29)
This function is plotted in Fig. 7.36. Had there been no inductance in the circuit, the current would have jumped immediately to \varepsilon _{0}/R. In practice, every circuit has some self-inductance, and the current approaches \varepsilon _{0}/R. asymptotically. The quantity τ ≡ L/R is the time constant; it tells you how long the current takes to reach a substantial fraction (roughly two-thirds) of its final value.
^{17} Notice that −L(d I /dt) goes on the left side of the equation—it is part of the emf that establishes the voltage across the resistor.
